Floor inequality: $\lfloor x+y\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$

Question

I remember seeing the inequality $\lfloor x+y\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor$ somewhere which is true for all reals.

So I was wondering what's wrong with this proof?

For all reals $a,b$ with $|b|>1$ we can use the inequality to get:

$\lfloor \frac{a-1}{b}\rfloor\ge\lfloor\frac{a}{b}\rfloor+\lfloor-\frac{1}{b}\rfloor=\lfloor\frac{a}{b}\rfloor$,

since $\lfloor-\frac{1}{b}\rfloor=1$ since $|b|>1$.

Is there anything wrong?

Answer 1

$$ \left\lfloor -\dfrac{1}{b} \right\rfloor = -1 $$ so you should get $$ \left\lfloor \dfrac{a-1}{b} \right\rfloor \ge \left\lfloor \dfrac{a}{b} \right\rfloor + \left\lfloor \dfrac{-1}{b} \right\rfloor = \left\lfloor \dfrac{a}{b} \right\rfloor - 1 $$